Download PHYS420 (Spring 2002) Riq Parra Homework # 7

PHYS420 (Spring 2002)
Riq Parra
Homework # 7 Solutions
Problems
By the end of this homework, chances are, you will be tired of computing the numerical
value for the product hc. Here’s a little suggestion-- these problems are simplified if we
express the combination of hc in units of eV ⋅ nm . That way you can work in convenient
units of eV and nm: hc = 1.986 x10 −25 J ⋅ m = 1239.7 eV ⋅ nm .
1.
In the typical experimental photoelectric set-up, a Geissler tube (two electrodes in
a evacuated glass tube) is connected to some voltage source. In this context, state
the difference between the stopping potential, a retarding voltage and an
accelerating voltage. Feel free to draw diagrams/graphs to make your point clear.
When a piece of metal is irradiated with UV light, electrons are ejected from the
surface. If the metal is connected to one end of a potential difference, the
electrons will see an electric field and feel an electric force. The electric field
lines are oriented so that they start on the positive plate and end on the negative
plat; in other words the electric force on the electron is always directed towards
the positive plate.
Accelerating voltage: If the voltage is set up so that the photoelectrons (i.e. those
ejected from the action of the UV light) come off the negative plate, they will feel
an accelerating voltage towards the positive plate.
Retarding voltage: If the voltage is reversed so that the photoelectrons come off
the positive plate the electrons will feel a decelerating voltage towards the
negative plate. The electrons will have a range of velocities (KE) when they are
ejected from the surface so, in general, slow electrons will not make it to the
negative plate but fast ones will.
Stopping potential: If we increase the retarding voltage so that not even the
fastest electrons make it, then the photocurrent stops. This is called the stopping
potential. Even those electrons emitted with the highest speeds just fail to reach
the plates.
2.
State and clearly explain three classical (i.e. Maxwell model for light) predictions
regarding the photoelectric effect. Feel free to draw diagrams/graphs to make your
point clear.
(1) The maximum kinetic energy should be proportional to the intensity of the
radiation.
(2) The photoelectric effect should occur for light of any frequency.
(3) There’s a measurable time delay between the radiation first striking the
surface and the first electrons being emitted.
(For a full detail explanation of each one, please refer to the book)
3.
A laser beam with an intensity of 120 W/m2 (roughly that of a small helium-neon
laser) is incident on a surface of sodium. It takes a minimum energy of 2.3 eV to
release an electron from sodium (the work function Φ of sodium). Assuming the
electron to be confined to an area of radius equal to that of a sodium atom (0.10
nm), how long will it take, classically, for the surface to absorb enough energy to
release an electron?
Classically, the average power Pav delivered by the waves of intensity I to an area
A is IA. An atom on the surface displays a ‘target area’ of A = π r2 = π (0.10x10-9
m)2 = 3.1x10-20 m2. If the entire electromagnetic power is delivered to the
electron, energy is absorbed at the rate ∆E/∆t = Pav. The time interval ∆t
necessary to absorb an energy ∆E = Φ can be expressed as
∆E Φ
(2.3eV )(1.6 x10 −19 J / eV )
∆t =
=
=
= 0.10 s
Pav IA (120 W / m 2 )(3.1x10 − 20 m 2 )
4.
State and clearly explain three experimental facts about the photoelectric effect.
Feel free to draw diagrams/graphs to make your point clear.
(1) The maximum kinetic energy (determined from the stopping potential) is
totally independent of the intensity of the light source.
(2) The maximum kinetic energy is a function of the frequency of light.
(3) The photoelectric effect does not occur at all if the frequency of the light
source is below a certain value.
(4) The first photoelectrons are emitted virtually instantaneously (within 10-9 s)
after the light source is turned on.
(For a full detail explanation of each one, please refer to the book)
5.
(a) What are the energy and momentum of a photon of red light of wavelength
650 nm? (b) What is the wavelength of a photon of energy 2.40 eV? Recall from
relativity that for a massless particle, E = pc .
(6.63x10 −34 J ⋅ s )(3.00 x10 8 m / s )
= 3.06 x10 −19 J = 1.91 eV.
λ
650 x10 −9 m
E h (6.63 x10 −34 J ⋅ s )
= 1.02 x10 − 27 kg ⋅ m / s = 1.91 eV/c.
(b) p = = =
−9
c λ
650 x10 m
(a) E = hf =
6.
hc
=
The work function for tungsten metal is 4.52 eV. (a) What is the cutoff
wavelength λc for tungsten? (b) What is the maximum kinetic energy of the
electrons when radiation of wavelength 198 nm is used? (c) What is the stopping
potential in this case?
(a) Starting from Einstein’s photoelectric equation KE max = hf − φ , the cutoff
frequency is when the electron has received just enough energy to overcome the
binding energy (a.k.a. the work function). Mathematically,
hc
0 = hf c − φ =
−φ
λc
(6.63x10−34 J ⋅ s )(3x108 m / s )
= 275 nm (in the UV)
φ
(4.52 eV )(1.60 x10−19 J / eV )
(b) The maximum kinetic energy is just given by,
hc
KE max = hf − φ =
−φ
λc =
hc
=
−34
λ
(6.63x10 J ⋅ s )(3x10 8 m / s )
− 4.52 eV (1.602 x10 −19 J / eV ) = 1.75 eV.
−9
198 x10 m
(c) The stopping potential is the voltage corresponding to KEmax:
KE max 1.75 eV
Vstopping =
=
= 1.75 Volts.
e
e
=
7.
Extra credit problem 1. SMM, Chapter 2, Problem 19.
We can solve this problem numerically from the very beginning, but we’ll do it
analytically first and then plug in at the last moment.
hc
For λ1, KE1 max = 1.00 eV, we have KE1 max =
−φ .
λ1
For λ2 = λ1/2, KE2 max = 4.00 eV, we have KE 2 max =
hc
−φ .
(λ1 / 2)
Solving one of them for the work function and substituting into the other one, we
get
hc
KE 2 max − KE1 max =
λ1
We can substitute this expression for hc/λ1 in the first equation and solve for the
work function.
φ = KE 2 max − 2 KE1 max = 4.00 eV – 2.00 eV = 2.00 eV
8.
Extra credit problem 2. Consider a photon with wavelength λ incident on an
unknown metal. The most energetic electrons ejected from the metal are bent into
a circular orbit of radius R by a magnetic field whose strength is equal to B. (a)
Derive an expression for the work function in terms of these variables. (b) Use
this expression to solve SMM, Chapter 2, Problem 21.
(a) A charged particle, with charge q and velocity v, traveling perpendicular to a
uniform magnetic field B will trace out a circular arc of R where the magnetic
v2
force acts like a centripetal force: FB = qvB = m . Solving for the velocity,
R
qBR
v=
m
Using Einstein’s photoelectric equation and solving for the work function,
φ = hf − KE max
hc 1
φ = − mv 2
λ 2
Substituting for the velocity in terms of B, R, q and m
2
1  qBR 
− m

λ 2  m 
Simplifying and realizing that we are using electrons,
hc e 2 B 2 R 2
φ= −
λ
2me
φ=
hc
(b) We can now use this equation to do SMM, Chapter 2, Problem 21. Plugging
all the parameters as given,
φ=
1239.7 eV ⋅ nm (1.602 x10 −19 C ) 2 (2.0 x10 −5 T ) 2 (0.20m) 2
1
−
−31
450 nm
2(9.11x10 kg )
1.602 x10 −19 J / eV
φ = 2.755 eV − 1.407 eV = 1.348 eV.